our only way (up to this point) to differentiate the expression is to expand it and get a polynomial, and then differentiate that polynomial. This method becomes very complicated and is particularly error prone when doing calculations by hand. A beginner might guess that the derivative of a product is the product of the derivatives, similar to the sum and difference rules, but this is not true. To take the derivative of a product, we use the product rule.

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Suppose one wants to differentiate ƒ(x) = x2 sin(x). By using the product rule, one gets the derivative ƒ '(x) = 2x sin(x) + x2cos(x) (since the derivative of x2 is 2x and the derivative of sin(x) is cos(x)).

One special case of the product rule is the constant multiple rule which states: if c is a real number and ƒ(x) is a differentiable function, then cƒ(x) is also differentiable, and its derivative is (c × ƒ)'(x) = c × ƒ '(x). This follows from the product rule since the derivative of any constant is zero. This, combined with the sum rule for derivatives, shows that differentiation is linear.

The rule for integration by parts is derived from the product rule, as is (a weak version of) the quotient rule. (It is a "weak" version in that it does not prove that the quotient is differentiable, but only says what its derivative is if it is differentiable.)

which is what you would get if you assumed the derivative of a product is the product of the derivatives.

To apply the product rule we multiply the first function by the derivative of the second and add to that the derivative of first function multiply by the second function. Sometimes it helps to remember the memorize the phrase "First times the derivative of the second plus the second times the derivative of the first."

The product rule can be used to give a proof of the power rule for whole numbers.
The proof proceeds by mathematical induction. We begin with the base case n=1{\displaystyle n=1}. If f1(x)=x{\displaystyle f_{1}(x)=x} then from the definition is easy to see that

Next we suppose that for fixed value of N{\displaystyle N}, we know that for fN(x)=xN{\displaystyle f_{N}(x)=x^{N}}, fN′(x)=NxN−1{\displaystyle f_{N}'(x)=Nx^{N-1}}. Consider the derivative of fN+1(x)=xN+1{\displaystyle f_{N+1}(x)=x^{N+1}},

We have shown that the statement fn′(x)=n⋅xn−1{\displaystyle f_{n}'(x)=n\cdot x^{n-1}} is true for n=1{\displaystyle n=1} and that if this statement holds for n=N{\displaystyle n=N}, then it also holds for n=N+1{\displaystyle n=N+1}. Thus by the principle of mathematical induction, the statement must hold for n=1,2,…{\displaystyle n=1,2,\dots }.